Unoccupied seats on flights cause airlines to lose revenue. Suppose a large airl

Question

Unoccupied seats on flights cause airlines to lose revenue. Suppose a large airline wants to estimate its average number of unoccupied seats per flight over the past year. To accomplish this, the records of 250 flights are randomly selected, and the number of unoccupied seats is noted for each of the sampled flights. The results are the following. Sample Size = 250

Sample Mean = 13.5956

Sample Standard Deviation = 5.1026

A. Compute the 95% confidence interval for the mean number of unoccupied seats.

B. Interpret the 95% confidence interval.

C. How many flights should be tested if the airline wants to be 95% confident of being within 1 seat of the population mean number of unoccupied seats?

D. How many flights should be tested if the airline wants to be 99% confident of being within 2 seats of the population mean number of unoccupied seats?

A food-products company conducted a market study by randomly sampling and interviewing 1,000 consumers to determine which brand of breakfast cereal they prefer. Suppose 424 consumers were found to prefer the company’s brand. Estimate the true fraction of all consumers who prefer the company’s cereal brand?

A. Compute the 95% confidence interval for the percent of consumers who prefer the company’s brand of breakfast cereal.

B. Interpret this confidence interval.

C. How large a sample size will need to be selected if we wish to have a 95% confidence interval that is accurate to within 1.5%. (Assume p = .40)

D. How large will the sample size need to be if we wish to be accurate to within 2.0%, with 95% confidence?

Explanation / Answer

A. here mean=13.5956, n=250 and sd=5.1026

z value is 1.96

now we will calculate ME=z*sd/sqrt(n)=1.96*5.1026/sqrt(250)=0.6325

CI=mean+/-ME=(12.9631,14.2281)

B. population mean will lie in this interval

C. here z=1.96, ME=1

so n=(z*sd/ME)^2=100

D. here z=2.576

n=(z*sd/ME)^2=43

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